Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x+y &= -4 \\ 6x-5y &= 8\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-5y = -6x+8$ Divide both sides by $-5$ to isolate $y$ $y = {\dfrac{6}{5}x - \dfrac{8}{5}}$ Substitute this expression for $y$ in the first equation. $3x+({\dfrac{6}{5}x - \dfrac{8}{5}}) = -4$ $3x + \dfrac{6}{5}x - \dfrac{8}{5} = -4$ Simplify by combining terms, then solve for $x$ $\dfrac{21}{5}x - \dfrac{8}{5} = -4$ $\dfrac{21}{5}x = -\dfrac{12}{5}$ $x = -\dfrac{4}{7}$ Substitute $-\dfrac{4}{7}$ for $x$ back into the top equation. $3( -\dfrac{4}{7})+y = -4$ $-\dfrac{12}{7}+y = -4$ $y = -\dfrac{16}{7}$ $y = -\dfrac{16}{7}$ The solution is $\enspace x = -\dfrac{4}{7}, \enspace y = -\dfrac{16}{7}$.